I have a project being setup in my garden for an aeroponics system its basic components are: 1. Esp8266 2. Single channel relay board 3. Temperature & humidity sensor 4. 12v 3a - 100psi pump 5. 2 x 12v 12ah batteries

I now want to add a charger and solar panel to maintain the power on this system or atleast reduce charge frequency as much as practically possible.

My setup will be 30 seconds every 2 minutes the pump turns on, the pump is 36 watts I'm overestimating the rest of the system to say 2.5 watts so we are possibly looking at using 38.5 watts for 30 seconds every 2 minutes.

Now a 40 watt solar panel is incredibly expensive for what I'm using it for and I have looked at the 4.5 watt version but how do I work out the correct panel I need.

Is there a proper way to do this? I just calculated it will have the pump running for 3.4 hours in a 24 hour period.

  • \$\begingroup\$ I dont really understand your pump numbers. You've said the pump goes on for 30 seconds every 2 minutes, but also that the pump is only running for an hour every day? \$\endgroup\$ – BeB00 Apr 12 '18 at 22:34
  • \$\begingroup\$ In total there are 1440 minutes in a day 2.33 minutes is a single cycle (2 minutes off 0.33 minutes on) so we get 618 cycles per day, if we calculate how many times the pump will be on we do 0.33 (time pump is on) x 618 (cycles per day) we get 204 minutes per day or 3.4 hours. \$\endgroup\$ – Chris James Apr 12 '18 at 22:40
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    \$\begingroup\$ Just edited there was a math error there sorry. \$\endgroup\$ – Chris James Apr 12 '18 at 22:41
  • \$\begingroup\$ 1/4 duty cycle of say 40W including 260-280W start means you need a capacity for run time between solar peaks with a 100W PV $170 to charge up and run in 4 h or run off the charger all day and forget the PV. Use a car battery and 4A float charger. You can do the math for ROI \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 12 '18 at 23:15
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    \$\begingroup\$ It is ESSENTIAL to have an accurate estimate of the power needed. So please measure the actual power using some type of logging ammeter product or something. Put your whole setup together except for the solar and charger, and run it off of a battery for 24 hours, logging the current of the battery over that time period. Once you know this, you can buy an appropriate charger and panels. I have a feeling you are using the rated current of the motor. But it may not use the full rated current depending on pump type and pressure head, etc. \$\endgroup\$ – mkeith Apr 13 '18 at 3:01

It's an energy equation.

38.5Watts, 3.4 hours is equivalent to 130.9Wh - watt hours - unit of energy.

To be net energy zero, you need at least 130.9Wh of energy per day. Now this depends a lot in your location but a good start is that you get 1000 effective hours of sun per year.

With 130.9Wh per day, you have 47.8kWh per year. If you have 1000 effective hours this would mean a solar panel of at least 47.8W.

Then it's more complicated because in winter you will have less, summer more, etc, etc.. but you get an initial approximation at least of what you would need.

Sizing the battery then also depends on location, weather, etc. That's another point altogether.

Hope this helps as a starting point. It doesn't consider efficiency in your converters, MPPT, charge/discharge efficiency, etc. All this could be grouped into a "safety coefficient"

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    \$\begingroup\$ Also, remember that a 40W solar panel means that the maximum output is 40W. You're basically never going to get that 40W for even 8 hours per day unless you're in the middle of the texas desert (nrel.gov/gis/solar.html) and your solar panel is kept at 25C, so you need a >300W cooling system \$\endgroup\$ – BeB00 Apr 13 '18 at 0:19
  • \$\begingroup\$ Thanks for everyone's answers I can now see how complicated this can get but know how to get myself a ball park figure, I will build the setup first then test by logging measurements. \$\endgroup\$ – Chris James Apr 13 '18 at 7:57
  • \$\begingroup\$ @ChrisJames, that's a good idea. I don't know if this is a hobby for you or what. But what I suggest, start with the calculations of what you expect to happen, build it, measure it, and see where the differences are and why is that happening. Basically, doing engineering. Then you can account for that the second time :D \$\endgroup\$ – Andrés Apr 13 '18 at 13:06
  • \$\begingroup\$ I agree with @Andrés that a preliminary calculation is a good idea. The calculate/measure/reconcile process can be carried out over several iterations. This is what leads to confidence that the system is well-behaved and well-understood. \$\endgroup\$ – mkeith Apr 13 '18 at 16:46

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